Test: Chemistry Minor Mock Test- 1 (March 16) - NEET MCQ

Is → As it is the ground state

Electron rich species are called lewis base. Among the given, BF₃ is an electron deficient species, so have a capacity of electrons accepting instead of donating that's why it is least likely to act as a lewis base. It is a lewis acid. 

Intensive means that the property is mass independent. We can see that Boiling point, pH, EMF of cell and surface tension are intensive pont . 
While Extensive property  depends on mass and in given options Volume and entropy are mass dependents.

Metalloids shows both the properties of metals as well as non metals.

⇒ H₂O is polar molecule
⇒ Hence  H²O has higher critical temp

Among the alkali metals cesium is the most reactive because the outermost electron is more loosely bound than the outermost electron of the other alkali metals.

Moles of CO =8 moles Moles of Fe₂O₃= 2.5 moles.

3 moles of CO is needed for 1 mole of Fe₂O₃ so 8 moles of CO will require 2.66 mole of Fe₂O₃ so Fe₂O₃ is limiting reagent.

1 mole of Fe₂O₃ produce 2 mole of Fe so 2.5 mole of Fe₂O₃will produce 5 mole of Fe = 280g of Fe.

Also 1 mole of Fe₂O₃ also produce 3 mole of CO₂ so 2.5 mole of Fe₂O₃ will produce 7.5 mole of CO₂=330g.

group is 15 so total valence electrons will be 5 while period is 3rd so n value of valence shell will be 3.

Balmer means transition to n = 2
1. line → 3 to 2
2. line → 4 to 2
3. line → 5 to 2

In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,

Where Ea = activation energy of forwarding reaction

Ea' = activation energy of backwards reaction

ΔH = enthalpy of the reaction

From the above diagram,

Ea = Ea' + ΔH

Thus, Ea > ΔH

Use ∆H=∆U+∆NgRT FOR 1. we need to find work done in constant volume that is ∆U which will be 0 as it is nCv∆T and since ∆T is zero so ∆U is zero .
For 2. we need to Find ∆H =∆U+∆NgRT Since ∆U is zero so∆H=∆NgRT 
R=2 T =300and ∆Ng is = -1

C-0/100-0 = F-32/180.
C/5= F-32/9.
C/5= 100-32/9.
C/5= 68/9.
C= 68×5/9.
C= 340/9.
C= 37.77.
C= 37.8

In this process, one molecule of hydrogen gas dissociated into two gaseous atoms i.e. the number of gaseous particles increase leading to  a more disordered state. 
.Breaking of bond requires energy

It is not mentioned that the process is isobaric. If we assume that process is isobaric because volume can not remain constant as number of gaseous particles are increased, then change in internal energy = 436 kJ mol^-1. The data of absorption gives us the value of q.

A photon is the quantum of electromagnetic radiation. The term quantum is the smallest elemental unit of a quantity, or the smallest discrete amount of something. Thus, one quantum of electromagnetic energy is called a photon. The plural of quantum is quanta.

for maximum ionic character,
* IE of cation must be less
* EA of anion must be larger
⇒ CS and  F
⇒ (B)

Let atomic weight of x = M_x
atomic weight of y = M_y 

we know, 
mole = weight /atomic weight 
a/c to question, 

mole of xy₂ = 0.1 
so, 
0.1 = 10g/( M_x +2M_y) 

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05 

0.05 = 9/( 3M_x + 2M_y ) 

3M_x + 2M_y = 9/0.05 = 9 × 20 = 180 g ---(2) 
solve eqns (1) and (2)

2M_x = 80 
M_x = 40g/mol 

and M_y = 30g/mole

Molar mass = 6(12) + 10(1) + 5(16) = 162g

Bronsted lowry acids are those acids which donate H⁺. In the reaction, A and D are giving H⁺. So, these both are bronsted lowry acid.

An electron bound to the nucleus by the mutual attraction between unlike charges, has a negative energy. Its most tightly bound to the nucleus nearer it is to the nucleus. That is an electron has least energy when it is in the n=1 shell. As the electron moves away from the nucleus its energy becomes less and less negative, i.e. its energy increases. When the electron is not under the attractive Coulomb force of the nucleus its potential energy is zero and whatever energy it has now, is kinetic energy, which is always positive. So farther the electron more energy it has.

For a buffer, we need to have a weak base/acid and its salt. Here, we have a weak base and so we will need its salt. In option c, concentration of strong acid is less than base. So we will have both a weak base and its salt. Therefore, only option c gives us a buffer solution.

Work done = - (change in moles) RT For work done positive change in moles negative and vice versa We only cosider the moles that are in gass phase.

Horizontal rows are periods while vertical column are groups.

As the size increases bond length also increases.

B₂H₆ is an electron deficient species. Soit will act as Lewis acid.

Orbital Angular momentum is given by [√l(l+1)]×h/2π
For s orbital, we have l = 0
Therefore orbital angular momentum = 0

Applying ideal gas eqn. at 1
p×22.4 = 1×0.0821×298        (since vol. Is in L. R should haave constaant with lit in it.)
p = 1.09 atm. To convert it into bar, we multiply it by 1.01 or we have 1.03 bar
Similarly at 2 and 3, we get value of pressure at 2 and 3. 
The most closest answer is option d

M, Kg, A are SI units.